HA YES I DID IT. HAHAHAHAHA :rofl:
I assumed E was the combined voltage sources and got -37.19 + 8i as the complex form
R + jwL = I/E
R +10iL = 96.2 + 66.38i
I had two variables and one equation, but I saw that there was real and nonreal numbers. So..
i terms -> 10iL = 66.38i
L = 6.638...
I'm not sure what "E" is meant to be, but I can meet you halfway I think :biggrin:
I = E /(R + jwL)
.3255cos(10t+133.3°) = E / [R + j(10)(L)]
-.223 + .237j = E / (R + 10jL)
If I were to take a guess at what 'E' is, it's the combination of the voltage sources, correct?
E = -7∠208° + 33∠-20° ?
Hmm.. Okay this took me a hell of a long time to calculate between those two equations. Was I supposed to use the equation in my first post and the equation found doing the KVL across L,R, and the two voltages to find R and L?
If so, I got these monstrous values (10^16) that have complex...
Are you telling me that the capacitor and 22Ω resistor might as well be non existent when solving for the voltage across R and L?
If so.. Does that mean I can do KVL while taking only R, L, and the 2 voltage sources into consideration? Giving me my 2nd equation?
Homework Statement
When the circuit shown in the figure is at steady state, the mesh current is
i(t) = 0.3255 cos(10t + 133.3°) A
Determine the values of L and R.
Zc is the impedance of the Capacitor
ZL is the impedance of the Inductor
I1 is the current in the top loop
I2 is the current in...
KCL on the top node:
0 = (Leaving currents) - (Entering Currents)
0 = (Current across Req) + (Current Source) - (Current through voltage source)
Current through voltage source = Isc
0= 1.237A + 9A - Isc
Isc = 10.237 A
The website won't tell me if this is correct (I used all my...
Ahh. Thank you for taking the time to draw that out.
Sorry, I'm getting a little confused from the different terminology. It sounds like you're asking what the voltage across the short circuit is. I'm not sure I know how to answer that. I'll take guesses though.
I'm supposing that the...
What he said.
A right triangle is basically like half of a square, cut down in a diagonal. The area of a square is b*h (or side^2). So a right triangle's area should be bh/2.
In the case of non-right triangles, this still holds true. Cut a parallelogram diagonally and you get two triangles...
Does it short out the 65 Ω resistor? If it does, I had no idea, because I thought for it to short out a parallel resistor, the short circuit has to be parallel ONLY to that resistor.
As for the potential across Req.. I guess I can use voltage division?
Veq= [8.889Ω/(8.889Ω+65Ω)]11v, which...
Homework Statement
For t<0, the circuit shown below with the voltage source at 11 volts and three resistors combined into an equivalent, Req. We need to find the Thevenin equivalent circuit for the part of the circuit connected to the inductor.
First find Req, by combining series and...
Homework Statement
This is a lab question.
Basically what the lab entails is we have two spheres. One is suspended from a string and another is attached to a non-conducting rod.
We inductively charge the sphere attached to the rod. Then we bring it close to the other sphere that is being...
Homework Statement
The springs BA and BC each have a stiffness of 500N/m and an unstretched length of 3m. Determine the displacement d of the cord from the wall when a force F = 175N is applied to the cord.
Homework Equations
F=ks
ƩFx=0
ƩFy=0
The Attempt at a Solution
This is as...